3.22 \(\int \frac{(a+b \tanh ^{-1}(c x))^2}{x^4} \, dx\)

Optimal. Leaf size=130 \[ -\frac{1}{3} b^2 c^3 \text{PolyLog}\left (2,\frac{2}{c x+1}-1\right )+\frac{1}{3} c^3 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{2}{3} b c^3 \log \left (2-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )-\frac{b c \left (a+b \tanh ^{-1}(c x)\right )}{3 x^2}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{3 x^3}-\frac{b^2 c^2}{3 x}+\frac{1}{3} b^2 c^3 \tanh ^{-1}(c x) \]

[Out]

-(b^2*c^2)/(3*x) + (b^2*c^3*ArcTanh[c*x])/3 - (b*c*(a + b*ArcTanh[c*x]))/(3*x^2) + (c^3*(a + b*ArcTanh[c*x])^2
)/3 - (a + b*ArcTanh[c*x])^2/(3*x^3) + (2*b*c^3*(a + b*ArcTanh[c*x])*Log[2 - 2/(1 + c*x)])/3 - (b^2*c^3*PolyLo
g[2, -1 + 2/(1 + c*x)])/3

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Rubi [A]  time = 0.230829, antiderivative size = 130, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5916, 5982, 325, 206, 5988, 5932, 2447} \[ -\frac{1}{3} b^2 c^3 \text{PolyLog}\left (2,\frac{2}{c x+1}-1\right )+\frac{1}{3} c^3 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{2}{3} b c^3 \log \left (2-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )-\frac{b c \left (a+b \tanh ^{-1}(c x)\right )}{3 x^2}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{3 x^3}-\frac{b^2 c^2}{3 x}+\frac{1}{3} b^2 c^3 \tanh ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x])^2/x^4,x]

[Out]

-(b^2*c^2)/(3*x) + (b^2*c^3*ArcTanh[c*x])/3 - (b*c*(a + b*ArcTanh[c*x]))/(3*x^2) + (c^3*(a + b*ArcTanh[c*x])^2
)/3 - (a + b*ArcTanh[c*x])^2/(3*x^3) + (2*b*c^3*(a + b*ArcTanh[c*x])*Log[2 - 2/(1 + c*x)])/3 - (b^2*c^3*PolyLo
g[2, -1 + 2/(1 + c*x)])/3

Rule 5916

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcT
anh[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcTanh[c*x])^(p - 1))/(1 -
c^2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 5982

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/d
, Int[(f*x)^m*(a + b*ArcTanh[c*x])^p, x], x] - Dist[e/(d*f^2), Int[((f*x)^(m + 2)*(a + b*ArcTanh[c*x])^p)/(d +
e*x^2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && LtQ[m, -1]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 5988

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^2)), x_Symbol] :> Simp[(a + b*ArcTanh[c
*x])^(p + 1)/(b*d*(p + 1)), x] + Dist[1/d, Int[(a + b*ArcTanh[c*x])^p/(x*(1 + c*x)), x], x] /; FreeQ[{a, b, c,
d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[p, 0]

Rule 5932

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_Symbol] :> Simp[((a + b*ArcTanh[c*
x])^p*Log[2 - 2/(1 + (e*x)/d)])/d, x] - Dist[(b*c*p)/d, Int[((a + b*ArcTanh[c*x])^(p - 1)*Log[2 - 2/(1 + (e*x)
/d)])/(1 - c^2*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 - e^2, 0]

Rule 2447

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[(Pq^m*(1 - u))/D[u, x]]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

Rubi steps

\begin{align*} \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{x^4} \, dx &=-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{3 x^3}+\frac{1}{3} (2 b c) \int \frac{a+b \tanh ^{-1}(c x)}{x^3 \left (1-c^2 x^2\right )} \, dx\\ &=-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{3 x^3}+\frac{1}{3} (2 b c) \int \frac{a+b \tanh ^{-1}(c x)}{x^3} \, dx+\frac{1}{3} \left (2 b c^3\right ) \int \frac{a+b \tanh ^{-1}(c x)}{x \left (1-c^2 x^2\right )} \, dx\\ &=-\frac{b c \left (a+b \tanh ^{-1}(c x)\right )}{3 x^2}+\frac{1}{3} c^3 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{3 x^3}+\frac{1}{3} \left (b^2 c^2\right ) \int \frac{1}{x^2 \left (1-c^2 x^2\right )} \, dx+\frac{1}{3} \left (2 b c^3\right ) \int \frac{a+b \tanh ^{-1}(c x)}{x (1+c x)} \, dx\\ &=-\frac{b^2 c^2}{3 x}-\frac{b c \left (a+b \tanh ^{-1}(c x)\right )}{3 x^2}+\frac{1}{3} c^3 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{3 x^3}+\frac{2}{3} b c^3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac{2}{1+c x}\right )+\frac{1}{3} \left (b^2 c^4\right ) \int \frac{1}{1-c^2 x^2} \, dx-\frac{1}{3} \left (2 b^2 c^4\right ) \int \frac{\log \left (2-\frac{2}{1+c x}\right )}{1-c^2 x^2} \, dx\\ &=-\frac{b^2 c^2}{3 x}+\frac{1}{3} b^2 c^3 \tanh ^{-1}(c x)-\frac{b c \left (a+b \tanh ^{-1}(c x)\right )}{3 x^2}+\frac{1}{3} c^3 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{3 x^3}+\frac{2}{3} b c^3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac{2}{1+c x}\right )-\frac{1}{3} b^2 c^3 \text{Li}_2\left (-1+\frac{2}{1+c x}\right )\\ \end{align*}

Mathematica [A]  time = 0.34386, size = 145, normalized size = 1.12 \[ -\frac{b^2 c^3 x^3 \text{PolyLog}\left (2,e^{-2 \tanh ^{-1}(c x)}\right )+a^2-2 a b c^3 x^3 \log (c x)+a b c^3 x^3 \log \left (1-c^2 x^2\right )+b \tanh ^{-1}(c x) \left (2 a-b c^3 x^3-2 b c^3 x^3 \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )+b c x\right )+a b c x+b^2 c^2 x^2+b^2 \left (1-c^3 x^3\right ) \tanh ^{-1}(c x)^2}{3 x^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcTanh[c*x])^2/x^4,x]

[Out]

-(a^2 + a*b*c*x + b^2*c^2*x^2 + b^2*(1 - c^3*x^3)*ArcTanh[c*x]^2 + b*ArcTanh[c*x]*(2*a + b*c*x - b*c^3*x^3 - 2
*b*c^3*x^3*Log[1 - E^(-2*ArcTanh[c*x])]) - 2*a*b*c^3*x^3*Log[c*x] + a*b*c^3*x^3*Log[1 - c^2*x^2] + b^2*c^3*x^3
*PolyLog[2, E^(-2*ArcTanh[c*x])])/(3*x^3)

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Maple [B]  time = 0.02, size = 339, normalized size = 2.6 \begin{align*} -{\frac{{a}^{2}}{3\,{x}^{3}}}-{\frac{{b}^{2} \left ({\it Artanh} \left ( cx \right ) \right ) ^{2}}{3\,{x}^{3}}}-{\frac{{c}^{3}{b}^{2}{\it Artanh} \left ( cx \right ) \ln \left ( cx-1 \right ) }{3}}-{\frac{c{b}^{2}{\it Artanh} \left ( cx \right ) }{3\,{x}^{2}}}+{\frac{2\,{c}^{3}{b}^{2}\ln \left ( cx \right ){\it Artanh} \left ( cx \right ) }{3}}-{\frac{{c}^{3}{b}^{2}{\it Artanh} \left ( cx \right ) \ln \left ( cx+1 \right ) }{3}}-{\frac{{b}^{2}{c}^{2}}{3\,x}}-{\frac{{c}^{3}{b}^{2}\ln \left ( cx-1 \right ) }{6}}+{\frac{{c}^{3}{b}^{2}\ln \left ( cx+1 \right ) }{6}}-{\frac{{c}^{3}{b}^{2} \left ( \ln \left ( cx-1 \right ) \right ) ^{2}}{12}}+{\frac{{c}^{3}{b}^{2}}{3}{\it dilog} \left ({\frac{1}{2}}+{\frac{cx}{2}} \right ) }+{\frac{{c}^{3}{b}^{2}\ln \left ( cx-1 \right ) }{6}\ln \left ({\frac{1}{2}}+{\frac{cx}{2}} \right ) }+{\frac{{c}^{3}{b}^{2}}{6}\ln \left ( -{\frac{cx}{2}}+{\frac{1}{2}} \right ) \ln \left ({\frac{1}{2}}+{\frac{cx}{2}} \right ) }-{\frac{{c}^{3}{b}^{2}\ln \left ( cx+1 \right ) }{6}\ln \left ( -{\frac{cx}{2}}+{\frac{1}{2}} \right ) }+{\frac{{c}^{3}{b}^{2} \left ( \ln \left ( cx+1 \right ) \right ) ^{2}}{12}}-{\frac{{c}^{3}{b}^{2}{\it dilog} \left ( cx \right ) }{3}}-{\frac{{c}^{3}{b}^{2}{\it dilog} \left ( cx+1 \right ) }{3}}-{\frac{{c}^{3}{b}^{2}\ln \left ( cx \right ) \ln \left ( cx+1 \right ) }{3}}-{\frac{2\,ab{\it Artanh} \left ( cx \right ) }{3\,{x}^{3}}}-{\frac{{c}^{3}ab\ln \left ( cx-1 \right ) }{3}}-{\frac{acb}{3\,{x}^{2}}}+{\frac{2\,{c}^{3}ab\ln \left ( cx \right ) }{3}}-{\frac{{c}^{3}ab\ln \left ( cx+1 \right ) }{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x))^2/x^4,x)

[Out]

-1/3*a^2/x^3-1/3*b^2/x^3*arctanh(c*x)^2-1/3*c^3*b^2*arctanh(c*x)*ln(c*x-1)-1/3*c*b^2*arctanh(c*x)/x^2+2/3*c^3*
b^2*ln(c*x)*arctanh(c*x)-1/3*c^3*b^2*arctanh(c*x)*ln(c*x+1)-1/3*b^2*c^2/x-1/6*c^3*b^2*ln(c*x-1)+1/6*c^3*b^2*ln
(c*x+1)-1/12*c^3*b^2*ln(c*x-1)^2+1/3*c^3*b^2*dilog(1/2+1/2*c*x)+1/6*c^3*b^2*ln(c*x-1)*ln(1/2+1/2*c*x)+1/6*c^3*
b^2*ln(-1/2*c*x+1/2)*ln(1/2+1/2*c*x)-1/6*c^3*b^2*ln(-1/2*c*x+1/2)*ln(c*x+1)+1/12*c^3*b^2*ln(c*x+1)^2-1/3*c^3*b
^2*dilog(c*x)-1/3*c^3*b^2*dilog(c*x+1)-1/3*c^3*b^2*ln(c*x)*ln(c*x+1)-2/3*a*b/x^3*arctanh(c*x)-1/3*c^3*a*b*ln(c
*x-1)-1/3*c*a*b/x^2+2/3*c^3*a*b*ln(c*x)-1/3*c^3*a*b*ln(c*x+1)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{3} \,{\left ({\left (c^{2} \log \left (c^{2} x^{2} - 1\right ) - c^{2} \log \left (x^{2}\right ) + \frac{1}{x^{2}}\right )} c + \frac{2 \, \operatorname{artanh}\left (c x\right )}{x^{3}}\right )} a b - \frac{1}{12} \, b^{2}{\left (\frac{\log \left (-c x + 1\right )^{2}}{x^{3}} + 3 \, \int -\frac{3 \,{\left (c x - 1\right )} \log \left (c x + 1\right )^{2} + 2 \,{\left (c x - 3 \,{\left (c x - 1\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )}{3 \,{\left (c x^{5} - x^{4}\right )}}\,{d x}\right )} - \frac{a^{2}}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))^2/x^4,x, algorithm="maxima")

[Out]

-1/3*((c^2*log(c^2*x^2 - 1) - c^2*log(x^2) + 1/x^2)*c + 2*arctanh(c*x)/x^3)*a*b - 1/12*b^2*(log(-c*x + 1)^2/x^
3 + 3*integrate(-1/3*(3*(c*x - 1)*log(c*x + 1)^2 + 2*(c*x - 3*(c*x - 1)*log(c*x + 1))*log(-c*x + 1))/(c*x^5 -
x^4), x)) - 1/3*a^2/x^3

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{2} \operatorname{artanh}\left (c x\right )^{2} + 2 \, a b \operatorname{artanh}\left (c x\right ) + a^{2}}{x^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))^2/x^4,x, algorithm="fricas")

[Out]

integral((b^2*arctanh(c*x)^2 + 2*a*b*arctanh(c*x) + a^2)/x^4, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \operatorname{atanh}{\left (c x \right )}\right )^{2}}{x^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x))**2/x**4,x)

[Out]

Integral((a + b*atanh(c*x))**2/x**4, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{artanh}\left (c x\right ) + a\right )}^{2}}{x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))^2/x^4,x, algorithm="giac")

[Out]

integrate((b*arctanh(c*x) + a)^2/x^4, x)